﻿ Perpendicular Lines Worksheet | Problems & Solutions

# Perpendicular Lines Worksheet

Perpendicular Lines Worksheet
• Page 1
1.
What is the slope of a line perpendicular to 2$x$ + 3$y$ + 9 = 0?
 a. $\frac{3}{2}$ b. - $\frac{3}{2}$ c. $\frac{2}{3}$ d. - $\frac{2}{3}$

#### Solution:

2x + 3y + 9 = 0 y = (- 2 / 3) x + (- 9 / 3) .
[Simplify.]

Slope of the line is - 2 / 3.
[Compare with y = mx + c.]

Slope of required line × Slope of 2x + 3y + 9 = 0 is - 1.
[Product of slopes of perpendicular lines is - 1.]

Slope of required line × - 2 / 3 = - 1
[Substitute.]

Slope of required line = 3 / 2
[Simplify.]

2.
What can you say about the lines 3$x$ + 2$y$ + 5 = 0 and 2$x$ - 3$y$ + 7 = 0?
 a. parallel b. perpendicular c. collinear d. at 60o with respect to each other

#### Solution:

Slope of 3x + 2y + 5 = 0 is - 3 / 2.
[Slope = - coefficient of xcoefficient of y .]

Slope of 2x - 3y + 7 = 0 is 2 / 3.
[Slope = - coefficient of xcoefficient of y .]

Product of slopes = - 3 / 2 × 2 / 3 = -1.

The given lines are perpendicular .
[If product of slopes is -1, then the lines are perpendicular.]

3.
If the line through ($x$, 3) and (4, 9) is perpendicular to the line through (- 1, 4) and (3, 6), then what is the value of $x$?
 a. 10 b. 16 c. 4 d. 7

#### Solution:

Let A(x, 3), B(4, 9), C(- 1, 4), D(3, 6) be the points.

Slope of AB = 9 - 34 - x = 64 - x
[Slope formula is m = y2 -y1x2 -x1.]

Slope of CD = 6 - 43 + 1 = 1 / 2.
[Slope formula.]

64 - x × 12 = - 1
[Product of slopes of perpendicular lines is - 1.]

34-x = - 1

x = 7.
[Simplify.]

4.
What is the product of the slopes of the lines $x$ = 7 and $y$ = 7?
 a. not defined b. 49 c. -1

#### Solution:

x = 3 is a vertical line.

Slope of a vertical line is not defined .

So, the product of the slopes of x = 3 and y = 3 is not defined .

5.
Identify the correct statement.
I. The coordinate axes are perpendicular to each other
II. The coordinate axes are vertical
III. The coordinate axes are horizontal.
IV. The coordinate axes are parallel to each other.
 a. II b. I c. III d. IV

#### Solution:

The coordinate plane is formed by two number lines, called the axes, intersecting at right angles.

The x-axis is the horizontal axis.

The y-axis is the vertical axis.

The two axes meet at the origin.

Hence the coordinate axes are perpendicular to each other.

6.
What is the slope of the line perpendicular to 2$x$ + 3$y$ + 6 = 0?
 a. $\frac{-2}{3}$ b. $\frac{2}{3}$ c. $\frac{3}{2}$ d. $\frac{-3}{2}$

#### Solution:

Slope of 2x + 3y + 6 = 0 is - 2 / 3.
[Slope = - coefficient of x coefficient of y )]

Slope of required line × Slope of 2x + 3y + 6 = 0 = -1.
[Product of slopes of perpendicular lines = -1.]

Slope of required line = 3 / 2
[Substitute and simplify.]

7.
A (- 3, 2), B (5, 1), C (2, 7), D (1, - 1) are the coordinates of 4 points in the $x$-$y$ plane. What can you say about the lines $\begin{array}{c}\stackrel{↔}{\text{AB}}\end{array}$ and $\begin{array}{c}\stackrel{↔}{\text{CD}}\end{array}$?
 a. at an angle of 45o with respect to each other b. perpendicular c. parallel d. same

#### Solution:

Slope of the line through (x1, y1) and (x2, y2) is y2 -y1x2 -x1.

Slope of AB = 1 - 25 - (-3) = - 1 / 8
[Step 1.]

Slope of CD = -1 - 71 - 2 = 8
[Step 1.]

Product of slopes of AB and CD = - (1 / 8) × 8 = - 1
[Step 2 and Step 3.]

Lines AB and CD are perpendicular.
[If the product of the slopes is - 1, then the lines are perpendicular.]

8.
What is the slope of a line perpendicular to $y$ = 4?
 a. not defined b. 4 c. -1

#### Solution:

As y = 4 is horizontal, any line perpendicular to it will be vertical.

Slope of required line is not defined .
[Slope of vertical lines is not defined.]

9.
Which of the following is not perpendicular to $y$ = 2$x$ + 6?
 a. 6$y$ + 3$x$ + 10 = 0 b. 2$y$ = $x$ + 5 c. 4$y$ + 2$x$ + 7 = 0 d. 8$x$ + 4$y$ + 8 = 0

#### Solution:

Slope of y = 2x + 6 is 2 .
[Compare with y = mx + b.]

2y = x + 5 y = 1 / 2x + 5 / 2.
[Rearrange.]

Slope of 2y = x + 5 is 1 / 2.
[Compare with y = mx + b.]

Product of slopes of y = 2x + 6 and 2y = x + 5 is 2 × 1 / 2 i.e. 1.
[Step 1 and Step 3.]

All other lines given except 2y = x + 5 are perpendicular to y = 2x + 6 .
[Slope of each is - 1 / 2.]

10.
Find the inclination of a line perpendicular to $y$ = $x$ - 3.
 a. 45o b. 60o c. 135o d. 90o

#### Solution:

Let the inclination of the required line be θ.

Slope of y = x - 3 is 1.
[Compare with y = mx + b.]

Slope of the required line is - 1.
[Product of slopes of perpendicular lines is - 1.]

tan θ = - 1
[Slope = tan θ, where θ is the inclination.]

θ = 135o
[tan 135o = - 1.]

The inclination of a line perpendicular to y = x - 3 is 135o.